Question: Solve for $x$ : $2\sqrt{x} + 9 = 10\sqrt{x} + 8$
Subtract $2\sqrt{x}$ from both sides: $(2\sqrt{x} + 9) - 2\sqrt{x} = (10\sqrt{x} + 8) - 2\sqrt{x}$ $9 = 8\sqrt{x} + 8$ Subtract $8$ from both sides: $9 - 8 = (8\sqrt{x} + 8) - 8$ $1 = 8\sqrt{x}$ Divide both sides by $8$ $\frac{1}{8} = \frac{8\sqrt{x}}{8}$ Simplify. $\dfrac{1}{8} = \sqrt{x}$ Square both sides. $\dfrac{1}{8} \cdot \dfrac{1}{8} = \sqrt{x} \cdot \sqrt{x}$ $x = \dfrac{1}{64}$